Physics - Modern Physics Question with Solution | TestHub

The transition equation for Lyman series is given by $\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$ for largest wavelength, $n=2$. $\therefore \frac{1}{{\lambda }_{\max }}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$ The transition equation for Balmer series is given by $\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$ for largest wavelength, $n=3$ $\therefore \frac{1}{{\lambda }_{\max }}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ Therefore, $\frac{{\lambda }_{L_{\max }}}{{\lambda }_{B_{\max }}}=\frac{\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{\left(\frac{1}{1^2}-\frac{1}{2^2}\right)}=\frac{\frac{5}{36}}{\frac{3}{4}}=\frac{5}{27}$